Question

A second pendulum clock having steel wire is calibrated at ${20}^{0}C$. When temperature is increased to ${30}^{0}C$, then how much time does the clock lose or gain in one week ?
$[{\alpha }_{steel}=1.2\times {10}^{-5}{{(}^{0}C)}^{-1}]$ :

• A. $0.3628s$
• B. $3.626s$
• C. $362.8s$
• D. $36.28S$

Answer 1

Answer: (D)

$36.28S$

As the wire is heated, it leads to the change in length due to change in temperature and also the change in length leads to the change in the time period of the pendulum.

The change in time period is given by:

$\frac{\Delta T}{T}=\frac{1}{2}\alpha \Delta \theta$

$\Delta T=\frac{1}{2}\alpha T\Delta \theta$

$T=1week=7\times 24\times 3600=604800sec$

$\Delta T=\frac{1}{2}\times (1.2\times {10}^{-5})\times 604800\times (30-20)$

$\Delta T=36.28sec$

So, the time lost by the clock in 1 week will be $36.28s$